26 is unique between a square (5^2) and a cube (3^3)

  1. i.e, [tex]a^{2} \pm 2 = b^{3}[/tex] has (25,27) as the only solution.

    Now, can you prove it?
     
  2. jcsd
  3. I think this is difficult problem. However, there is a trivial solution (1)^2-2=(-1)^3.
     
  4. Hi Robert, I'm sorry I didn't clarify that a, b must be positive integers.
     
  5. Are you asking for help in proving the statement or challenging us to give a proof?
     
  6. second option :)
     
  7. It's easy if you use Z[sqrt(-2)] and Z[sqrt(2)] (they're both UFD).

    For example look at a^2 + 2 = b^3. This factors as (a + sqrt(-2))(a - sqrt(-2)) = b^3, and the units of Z[sqrt(-2)] are just +/-1, so

    a + sqrt(-2) = (c + d*sqrt(-2))^3 for some c,d in Z.

    So

    a = c^3 - 6cd^2 = c*(c^2 - 6d^2) and
    1 = d*(3c^2 - 2d^2).

    So d = +/-1 and c = +/- 1, whence you get a = +/-5, and so b = 2.

    Same situation for a^2 - 2 = b^3.
     
  8. Seem O.K. to me, except b=3. In the second case we have a=+/-1 b=-1. However, Dr. Math makes it more complicated: http://mathforum.org/library/drmath/view/51569.html
     
  9. Yea, I don't know why the post in that link is using both a + sqrt(-2) = w^3 and a - sqrt(-2) = t^3 (conjugate). You only need the first, you get just as much information just more quickly.
     
  10. To go from

    [tex](a + \sqrt{-2})(a - \sqrt{-2}) = b^3[/tex]

    to

    [tex]a + \sqrt{-2} = (c + d\sqrt{-2})^3[/tex]

    don't we need that [itex]a + \sqrt{-2}[/itex] and [itex]a - \sqrt{-2}[/itex] are relatively prime? In fact, as Doctor Rob points out on the Math Forum, GCD([itex]a + \sqrt{-2}, a - \sqrt{-2}[/itex]) divides [itex]2\sqrt{-2}[/itex]. I think that's why he has to consider several cases.
     
  11. Yes, that's right. So what I wrote above is not entirely correct, but one gets the idea - just factor and use UFD-ness of Z[sqrt(-2)].

    More importantly, does anyone know how to use tex here without having to type itex and \itex everytime I want it? for example i prefer to $ $, just much easier.
     
  12. The pdf file linked at the beginning of this thread strongly suggests that itex, /itex and its variants are the only format that works here.
     
  13. Hi folks, I followed a different direction.

    first notice that a and b must have the same parity, so first case:

    1) a, b = even

    [tex](2k)^{2}\pm 2 = (2n)^{3}[/tex]
    [tex]4k^{2}\pm 2 = 8n^{3}[/tex]
    so,
    [tex]2k^{2}\pm 1= 4n^{3}[/tex]

    absurd, since even number cannot be equal to odd number, so second case:

    2) a, b = odd

    [tex](2k+1)^{2}\pm 2 = (2n+1)^{3}[/tex]

    2.1) + 2

    [tex]4k(k+1)+3 = 8n^{3}+12n^{2}+6n+1[/tex]

    wich means

    [tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod 4[/tex]
    [tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod k[/tex]
    and
    [tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod (K+1)[/tex]

    so

    [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod 4[/tex]
    [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod k[/tex]
    and
    [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod (K+1)[/tex]

    given [tex]f(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + .... + a_{i}x^{m_{i}}[/tex]
    and [tex]g(x) = b_{0} + b_{1}x + b_{2}x^{2} + b_{3}x^{3} + ....+ b_{j}x^{n_{j}}[/tex]

    we know that

    [tex]f(x)\equiv\ g(x)\mod W[/tex] [tex]\Leftrightarrow[/tex]
    [tex] a_{0}+a_{1}+a_{2}+a_{3}+ .... +a_{i}\equiv\ b_{0}+b_{1}+b_{2}+b_{3}+ ....+b_{j}\ mod W[/tex]

    lets make

    [tex]f(x) = 4n^{3}+6n^{2}+3n[/tex]

    and

    [tex]g(x) = 1[/tex]

    finally (skip obvious manipulations)

    [tex]12\equiv\ 0 \mod 4[/tex]
    [tex]12\equiv\ 0 \mod k[/tex]
    and
    [tex]12\equiv\ 0 \mod (K+1)[/tex]

    k must be = 2

    2.2) - 2

    (...) (after the same steps)

    finally

    [tex]14\equiv\ 0 \mod 4[/tex]

    what is untrue

    Any remarks would be apreciated.
     
  14. Is there anything not clear, or perhaps wrong? I apreciate some feedback
     
    Last edited: Nov 2, 2010
  15. Thinking again, I suspect I misunderstood the theorem found here:

    http://planetmath.org/encyclopedia/PolynomialCongruence.html

    and jumped to a conclusion prematurely.

    [tex]f(x)\equiv\ g(x)\mod W[/tex]
    [tex] a_{0}+a_{1}+a_{2}+a_{3}+ .... +a_{i}\equiv\ b_{0}+b_{1}+b_{2}+b_{3}+ ....+b_{j}\ mod W[/tex]

    Seems that the conclusion above is wrong.

    Anyway, I'd preciate if anyone could make the theorem more clear to me, and make any comments about what I've posted :)

    thank you!
     
  16. [tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod 4[/tex]
    [tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod k[/tex]
    and
    [tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod (k+1)[/tex]

    so

    [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod 4[/tex]
    [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod k[/tex]
    and
    [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod (k+1)[/tex]


    that's also wrong, but do not detroy the following argument because

    [tex]24\equiv\ 0 \mod 4[/tex]
    [tex]24\equiv\ 0 \mod k[/tex]
    and
    [tex]24\equiv\ 0 \mod (k+1)[/tex]

    still would imply that k must be = 2
     
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